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audioop Module Complexity¶
The audioop module operates on fragments of raw PCM audio: signed integer samples 1, 2, 3 or 4
bytes wide, held in any contiguous bytes-like object such as bytes, bytearray, memoryview or
array.array. Every function is C, reads the fragment in place without copying it, and returns
numbers or new bytes objects; nothing is kept between calls except the state tuple you
pass back to ratecv(), lin2adpcm() and adpcm2lin().
Removed in Python 3.13
Deprecated in Python 3.11 and removed in Python 3.13 by PEP 594. The examples need Python 3.11 or 3.12.
n is the bytes in a fragment (each fragment, where a function takes two of the same length),
m is the bytes in the reference passed to findfit(), and o is the bytes ratecv()
returns. The sample width is at most 4 bytes, so the work per sample is O(1), and an output that
is a fixed multiple of the input, such as ulaw2lin()'s n·width bytes, is O(n).
Complexity Reference¶
Analysis¶
| Operation | Time | Space | Notes |
|---|---|---|---|
audioop.getsample(fragment, width, index) |
O(1) | O(1) | Raises audioop.error for an index outside the fragment |
audioop.max(fragment, width) |
O(n) | O(1) | Largest absolute sample value |
audioop.minmax(fragment, width) |
O(n) | O(1) | Smallest and largest sample, in one pass |
audioop.avg(fragment, width) |
O(n) | O(1) | Mean of the samples |
audioop.rms(fragment, width) |
O(n) | O(1) | Root mean square, a measure of power |
audioop.cross(fragment, width) |
O(n) | O(1) | Number of zero crossings |
audioop.avgpp(fragment, width), audioop.maxpp(fragment, width) |
O(n) | O(1) | Average and largest peak-to-peak value |
Transformation¶
| Operation | Time | Space | Notes |
|---|---|---|---|
audioop.add(fragment1, fragment2, width) |
O(n) | O(n) | The fragments must be the same length; a sum that overflows is clipped |
audioop.mul(fragment, width, factor) |
O(n) | O(n) | A product that overflows is clipped |
audioop.bias(fragment, width, bias) |
O(n) | O(n) | A sum that overflows wraps around |
audioop.reverse(fragment, width) |
O(n) | O(n) | Reverses the sample order |
audioop.byteswap(fragment, width) |
O(n) | O(n) | Swaps the byte order of every sample |
audioop.tomono(fragment, width, lfactor, rfactor) |
O(n) | O(n) | Returns n/2 bytes |
audioop.tostereo(fragment, width, lfactor, rfactor) |
O(n) | O(n) | Returns 2n bytes |
audioop.lin2lin(fragment, width, newwidth) |
O(n) | O(n) | Returns n·newwidth/width bytes |
audioop.ratecv(fragment, width, nchannels, inrate, outrate, state, weightA=1, weightB=0) |
O(n + o) | O(o) | o is about n·outrate/inrate. Returns (data, state); passing the state to the next call makes chunked output identical to one call. The buffer is rounded up to a multiple of outrate/gcd(inrate, outrate) frames, so rates with a small gcd allocate up to that many extra frames per call |
Codecs¶
| Operation | Time | Space | Notes |
|---|---|---|---|
audioop.lin2ulaw(fragment, width), audioop.lin2alaw(fragment, width) |
O(n) | O(n) | One byte per sample: n/width bytes |
audioop.ulaw2lin(fragment, width), audioop.alaw2lin(fragment, width) |
O(n) | O(n) | One sample per byte: n·width bytes |
audioop.lin2adpcm(fragment, width, state) |
O(n) | O(n) | Four bits per sample: n/(2·width) bytes, rounded down. Returns (data, state); the state does not carry an odd sample's half byte |
audioop.adpcm2lin(fragment, width, state) |
O(n) | O(n) | Two samples per byte: 2·n·width bytes. Returns (data, state) |
Search¶
These three take 16-bit samples only, and have no width argument.
| Operation | Time | Space | Notes |
|---|---|---|---|
audioop.findfit(fragment, reference) |
O((n − m + 1)·m) | O(1) | Rescores the whole reference at every offset, so a reference half the fragment's length is the quadratic worst case. fragment must be at least as long. Returns (offset, factor) |
audioop.findfactor(fragment, reference) |
O(n) | O(1) | The fragments must be the same length; the factor for one fixed alignment |
audioop.findmax(fragment, length) |
O(n) | O(1) | Slides a running sum, so length does not change the cost |
Exceptions¶
| Operation | Time | Space | Notes |
|---|---|---|---|
audioop.error |
O(1) | O(1) | Raised for a width other than 1 to 4, a fragment that is not a whole number of samples, and mismatched lengths |
Measuring a Fragment¶
The analysis functions read the fragment where it is and return a number, so they cost one pass and no memory, whatever kind of buffer holds the samples.
import array
import audioop
samples = array.array('h', [0, 1000, -3000, 2000, -500, 0]) # 16-bit samples
fragment = samples.tobytes()
assert audioop.max(fragment, 2) == 3000 # O(n)
assert audioop.minmax(fragment, 2) == (-3000, 2000) # O(n) - both in one pass
assert audioop.rms(fragment, 2) == 1541 # O(n)
assert audioop.cross(fragment, 2) == 4 # O(n)
assert audioop.getsample(fragment, 2, 2) == -3000 # O(1)
# Any bytes-like object works, with no copy to bytes first
assert audioop.max(samples, 2) == 3000 # O(n), O(1) memory
assert audioop.max(memoryview(fragment)[4:], 2) == 3000
try:
audioop.max(fragment[:-1], 2) # 11 bytes is not a whole number of samples
except audioop.error as error:
assert 'whole number' in str(error)
else:
raise AssertionError('a partial sample was accepted')
Converting and Mixing¶
Each transformation returns a new fragment, so a chain of them allocates one output per step. The output's length depends on what the function changes: the channel count, the sample width or the encoding.
import array
import audioop
mono = array.array('h', [100, -200, 300, -400]).tobytes() # 4 samples, 8 bytes
louder = audioop.mul(mono, 2, 2.0) # O(n)
assert audioop.max(louder, 2) == 800
mixed = audioop.add(mono, louder, 2) # O(n) - same lengths required
assert audioop.getsample(mixed, 2, 3) == -1200
stereo = audioop.tostereo(mono, 2, 1, 1) # O(n), 2n bytes
assert len(stereo) == 16
assert audioop.tomono(stereo, 2, 0.5, 0.5) == mono # O(n), n/2 bytes
wide = audioop.lin2lin(mono, 2, 4) # O(n), n·newwidth/width bytes
assert len(wide) == 16 and audioop.getsample(wide, 4, 0) == 100 << 16
# Overflow clips in add() and mul(), but wraps in bias()
assert audioop.add(b'\x7f', b'\x7f', 1) == b'\x7f'
assert audioop.bias(b'\x7f', 1, 1) == b'\x80'
Resampling and Encoding in Chunks¶
ratecv(), lin2adpcm() and adpcm2lin() return a state with their output. Passing it to the
next call continues the stream where the last chunk stopped, so a long recording can be
converted a chunk at a time in memory proportional to the chunk, with the same result as one call
on the whole. For lin2adpcm() that holds when each chunk has an even number of samples: an odd
one's last half byte is dropped, not carried in the state.
import array
import audioop
signal = array.array('h', (i * 37 % 2000 - 1000 for i in range(4410))).tobytes()
whole, _ = audioop.ratecv(signal, 2, 1, 44100, 8000, None) # O(n + o)
state = None
pieces = []
for start in range(0, len(signal), 882):
chunk = signal[start:start + 882]
converted, state = audioop.ratecv(chunk, 2, 1, 44100, 8000, state) # O(chunk)
pieces.append(converted)
assert b''.join(pieces) == whole
assert len(whole) // 2 == 800 # o is about n·outrate/inrate
# The ADPCM coders carry their predictor the same way
encoded, adpcm_state = audioop.lin2adpcm(signal, 2, None) # O(n), n/4 bytes for 16-bit
assert len(encoded) == len(signal) // 4
decoded, _ = audioop.adpcm2lin(encoded, 2, None) # O(n), 4 bytes per input byte
assert len(decoded) == len(signal)
u-LAW and A-LAW¶
The logarithmic codecs store one byte per sample, so encoding divides the size by the width and decoding multiplies it.
import array
import audioop
linear = array.array('h', [0, 1000, -1000, 32000]).tobytes()
ulaw = audioop.lin2ulaw(linear, 2) # O(n), n/width bytes
assert len(ulaw) == 4
back = audioop.ulaw2lin(ulaw, 2) # O(n), n·width bytes
assert len(back) == len(linear)
assert abs(audioop.getsample(back, 2, 1) - 1000) < 50 # lossy, but close
alaw = audioop.lin2alaw(linear, 2) # O(n)
assert len(audioop.alaw2lin(alaw, 4)) == 16 # decode straight to 32-bit samples
Searching a Fragment¶
findmax() keeps a running sum as its window slides, so it is one pass whatever the window
length. findfit() recomputes the whole correlation with the reference at every offset; when the
alignment is already known, findfactor() gives the factor in linear time instead.
import array
import audioop
quiet = [i * 7 % 21 - 10 for i in range(100)]
loud = [3000, 1000, -2000, -500, 2500, -3000, 100, 0, -1500, 1200]
fragment = array.array('h', quiet + loud + quiet).tobytes() # 16-bit only
assert audioop.findmax(fragment, 10) == 100 # O(n), whatever the length
reference = array.array('h', loud).tobytes()
offset, factor = audioop.findfit(fragment, reference) # O((n - m + 1)·m)
assert offset == 100 and abs(factor - 1.0) < 1e-9
aligned = fragment[2 * offset:2 * offset + len(reference)]
assert audioop.findfactor(aligned, reference) == 1.0 # O(n), same lengths
try:
audioop.findfit(reference, fragment) # the reference must be the shorter
except audioop.error as error:
assert 'longer' in str(error)
else:
raise AssertionError('a longer reference was accepted')
Common Patterns¶
Measuring a WAV File¶
import array
import audioop
import io
import wave
buffer = io.BytesIO()
with wave.open(buffer, 'wb') as out:
out.setnchannels(2)
out.setsampwidth(2)
out.setframerate(8000)
out.writeframes(audioop.tostereo(array.array('h', [16, -16] * 400).tobytes(), 2, 1, 1))
buffer.seek(0)
with wave.open(buffer, 'rb') as wav:
width = wav.getsampwidth()
frames = wav.readframes(wav.getnframes()) # O(n)
mono = audioop.tomono(frames, width, 0.5, 0.5) # O(n)
assert audioop.rms(mono, width) == 16 # O(n)
assert audioop.cross(mono, width) == 799
Performance Best Practices¶
✅ Do:
- Pass
bytearray,memoryvieworarray.arrayfragments straight in; they are read in place - Carry the state returned by
ratecv(),lin2adpcm()andadpcm2lin()into the next call, and process long recordings a chunk at a time - Use
findfactor()when the alignment is known: linear instead of a pass per offset - Keep a
findfit()search small: it costs the offsets tried times the reference length
❌ Avoid:
- A
findfit()reference near half the fragment's length - that is its quadratic worst case - Long chains of transformations on a large fragment - each step allocates a full new output
audioopin new code - it does not exist from Python 3.13
Version Notes¶
- Python 3.11+: Importing the module emits a
DeprecationWarning - Python 3.13+: Removed by PEP 594;
import audioopraisesModuleNotFoundError